For a map $f(t_1,t_2)=\langle t_1^3,t_1^2t_2,t_1t_2^2,t_2^3\rangle$ prove that f is one-to-one

Question

For a map $f(t_1,t_2)=(t_1^3,t_1^2t_2,t_1t_2^2,t_2^3)$, prove that f is one-to-one

I tried taking $t_1^3=x, t_1^2t_2=p, t_1 t_2^2=q,t_2^3=y$ and rewriting the equation as $xy=pq$. But i don't know how to check this equation for One-to-one?

Answer 1

1) The map $f$ is not one-to-one in general.
For example over $\mathbb C$ if $j=\frac {-1+i\sqrt 3}{2}$ we have $(1,0)\neq (j,0)$ and nevertheless $f(1,0)=f(j,0)=(1,0,0,0)$

2) The map $f$ is one-to-one for certain fields, for example $\mathbb R.$
Indeed if we know that $f(t_1,t_2)=(x,p,q,y)$ we can compute $t_1,t_2$ by $t_1=\sqrt [3]x$ and $t_2=\sqrt [3]y$.
The same argument is valid for every field such that $x\mapsto x^3$ is one-to-one, in particular for all subfields of $\mathbb R$.

3) Algebraic geometry interpretation.
The map $f$ induces a map $\mathbb P^1\to \mathbb P^3:[t_1:t_2]\mapsto [t_1^3:t_1^2t_2:t_1t_2^2:t_2^3]$ which is one-to-one and whose image is the so-called rational normal cubic curve in projective $3$-space.
This curve is important because it is the simplest example of a curve which is not the complete intersection of two surfaces in $\mathbb P^3$.